Problem: The lifespans of lizards in a particular zoo are normally distributed. The average lizard lives $3.2$ years; the standard deviation is $0.7$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lizard living between $1.8$ and $4.6$ years.
Answer: $3.2$ $2.5$ $3.9$ $1.8$ $4.6$ $1.1$ $5.3$ $95\%$ We know the lifespans are normally distributed with an average lifespan of $3.2$ years. We know the standard deviation is $0.7$ years, so one standard deviation below the mean is $2.5$ years and one standard deviation above the mean is $3.9$ years. Two standard deviations below the mean is $1.8$ years and two standard deviations above the mean is $4.6$ years. Three standard deviations below the mean is $1.1$ years and three standard deviations above the mean is $5.3$ years. We are interested in the probability of a lizard living between $1.8$ and $4.6$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the lizards will have lifespans within 2 standard deviations of the average lifespan. The probability of a particular lizard living between $1.8$ and $4.6$ years is ${95\%}$.